2X +5=-11
e) ( x + 3)^3 = ( 2x) ^3
f) ( 5 - x )^5 = 32
g) ( 5x - 6)^3 = 64
h)5 . 9^x = 405
i) 11^5 : 11^ n - 2 = 11^5
k) ( 3x)^3 = ( 2x + 1)^3
e) \(\left(x+3\right)^3=\left(2x\right)^3\)
\(\Rightarrow x+3=2x\)
\(\Rightarrow2x-x=3\)
\(\Rightarrow x=3\)
f) \(\left(5-x\right)^5=32\)
\(\Rightarrow\left(5-x\right)^5=2^5\)
\(\Rightarrow5-x=2\)
\(\Rightarrow x=5-2\)
\(\Rightarrow x=3\)
g) \(\left(5x-6\right)^3=64\)
\(\Rightarrow\left(5x-6\right)^3=4^3\)
\(\Rightarrow5x-6=4\)
\(\Rightarrow5x=4+6\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=\dfrac{10}{2}\)
\(\Rightarrow x=5\)
h) \(5\cdot9^x=405\)
\(\Rightarrow9^x=\dfrac{405}{5}\)
\(\Rightarrow9^x=81\)
\(\Rightarrow9^x=9^2\)
\(\Rightarrow x=2\)
i) \(11^5:11^{n-2}=11^5\)
\(\Rightarrow11^{n-2}=11^5:11^5\)
\(\Rightarrow11^{n-2}=1\)
\(\Rightarrow11^{n-2}=11^0\)
\(\Rightarrow n-2=0\)
\(\Rightarrow n=2\)
k) \(\left(3x\right)^3=\left(2x+1\right)^3\)
\(\Rightarrow3x=2x+1\)
\(\Rightarrow3x-2x=1\)
\(\Rightarrow x=1\)
(2x-5).(4-3x)-(3x+11).(5-2x)-15(2x-5)
\(\left(2x-5\right).\left(4-3x\right)-\left(3x+11\right).\left(5-2x\right)-15\left(2x-5\right)\)
\(=\)\(\left(2x-5\right).\left(4-3x\right)+\left(3x+11\right).\left(2x-5\right)-15\left(2x-5\right)\)
\(=\left(2x-5\right).\left(4-3x+3x+11-15\right)\)
\(=\left(2x-5\right).0\)
\(=0\)
(2x-5).(4x-3x)-(3x+11).(5-2x)+5.(2x-5)
(2x-5).(4x-3x)-(3x+11).(5-2x)+5.(2x-5)
=(2x-5).x+(3x+11).(2x-5)+5(2x-5)
=(2x-5)(x+3x+11+5)
=(2x-5)(4x+16)
=8x2+32x-20x-80
=8x2+12x-80
1.a)(2x - 5 ). (4 -3x) - (3x +11) .(5-2x) -15 . (2x -5)
giải
(2x-5).(4-3x)-(3x+11).(5-2x)-15.(2x-5)
8x-6x^2-20+15x-(15x-6x^2+55-22x)-30x+75
8x-6x^2-20+15x-15x+6x^2-55+22x-30x+75
(8x+15x-15x+22x-30x)+(-6x^2+6x^2)+(-20-55+75)
0+0+0=0
chúc bạn học tốt nha
Tình hợp lý nếu có thể :
-5/2x 2/11+-5/7x 9/11+15/7
Tìm x biết:
(2x-15)mũ 5=(2x-15)mũ 3
25/7nha
Chứng minh rằng giá trị của biểu thức sau không phụ thuộc vào giá trị của x
b) B = 2x(x – 3) – (2x – 2)(x – 2)
c) C = (3x – 5)(2x + 11) – (2x – 2)(3x + 7)
d) D = (2x + 11)(3x – 5) – (2x + 3)(3x + 7
b: \(B=2x\left(x-3\right)-\left(2x-2\right)\left(x-2\right)\)
\(=2x^2-6x-2x^2+4x+2x-4\)
=-4
giải bài toán tìm x sau:(2x-11)^5 = (2x-11)^3 với
=> 2x-11 bằng 1 hoặc 0
=> x= 6 hoặc 11/2
[ ( 2x - 11 ) : 3 + 11 ] x 5 = 20
(2x-11):3+11=4
(2x-11):3=15
2x-11=5
2x=16
x=8
[(2x - 11) : 3 + 1] × 5 = 20
(2x - 11) : 3 + 1 = 20 : 5
(2x - 11) : 3 + 1 = 4
(2x - 11) : 3 = 4 - 1
(2x - 11) : 3 = 3
2x - 11 = 3 × 3
2x - 11 = 9
2x = 9 + 11
2x = 20
x = 20 : 2
x = 10
Vậy x = 10
~ Học tốt ~ .❤
tìm x biết:
a)(2x+2)(2x-2)-4x(x+5)=8
b)(4x+5)(4x-5)-8x(2x-7)=11
c)(1/2x-3)(1/2x+3)-1/4x(x+5)=11/2
d)(3x+2)(3x-2)-4x(x+2-5x2=18
\(a.x=-0,6\)
\(c.x=-11,6\)
Pt nhju ak!!!
22/25 x 5/11 - 33/25 : 11/5 + 58/11
8 3/11 - (3 5/7 - 7 8/11 )
1/5 + 7/5 : x = -3/20
5/6x + 325% = 3/2x -30,75
\(\frac{22}{25}\times\frac{5}{11}-\frac{33}{25}:\frac{11}{5}+\frac{58}{11}\)
\(=\frac{22}{25}\times\frac{5}{11}-\frac{33}{25}\times\frac{5}{11}+\frac{58}{11}\)
\(=\frac{5}{11}\times\left(\frac{22}{25}+\frac{33}{25}\right)+\frac{58}{11}\)
\(=\frac{5}{11}\times\frac{11}{5}+\frac{58}{11}\)
\(=1+\frac{58}{11}\)
\(=\frac{11}{11}+\frac{58}{11}\)
\(=\frac{69}{11}\)
\(8\frac{3}{11}-\left(3\frac{5}{7}-7\frac{8}{11}\right)\)
\(=\frac{91}{11}-\left(\frac{26}{7}-\frac{85}{11}\right)\)
\(=\frac{91}{11}-\frac{26}{7}+\frac{85}{11}\)
\(=\frac{91}{11}+\frac{85}{11}-\frac{26}{7}\)
\(=16-\frac{26}{7}\)
\(=\frac{86}{7}\)
\(\frac{5}{6}x+325\%=\frac{3}{2}x-30,75\)
\(=>\frac{5}{6}x+3,25=\frac{3}{2}x-30,75\)
\(=>\frac{5}{6}x-\frac{3}{2}x=-30,75-3,25\)
\(=>\frac{-2}{3}x=-34\)
\(=>x=-34:\left(-\frac{2}{3}\right)\)
\(=>x=-34.\left(\frac{3}{-2}\right)\)
\(=>x=51\)